The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the standard deviation of requests for nonsmoking rooms is 5.25 customers.

Question

Quizplus · Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials (in this case, 25 calls), and $p$ is the probability of success (in this case, 0.70 for requesting nonsmoking rooms). Plugging in these values, we get $\sigma = \sqrt{25 \cdot 0.70 \cdot (1-0.70)} = \sqrt{25 \cdot 0.70 \cdot 0.30} = \sqrt{5.25} \approx 2.29$, not 5.25.

The Nationwide Motel Company Has Determined That 70 Percent of All